Quantitative Aptitude - Arithmetic Ability Questions

I. gives, S.I for 3 years = Rs. 4500.

II. gives, Rate = 10% p.a.

III. gives, (C.I.) - (S.I.) = Rs. 465.

Clearly, using I and III we get C.I. = Rs. (465 + 4500).

Thus, II is redundant.

Also, from I and II, we get sum =

100 x 4500

= 15000.

10 x 3

Now C.I. on Rs. 15000 at 10% p.a. for 3 years may be obtained.

Thus, III is redundant.

Either II or III is redundant.

3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7

5.25 meters=525 cm = 5 × 5 × 3 × 7

Hence common factors are 3 and 7

Hence LCM = 3 × 7 = 21

Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.

1 x 1 + 1 x 1 = (1 + 1) = 2

2 x 2 + 2 x 2 = (4 + 4) = 8

8 x 3 + 3 x 3 = (24 + 9) = 33

33 x 4 + 4 x 4 = (132 + 16) = 148

148 x 5 + 5 x 5 = (740 + 25) = 765.

Let the two conditioners be A and B

'A' cools at 40min

'B' at 45min

Together = (a x b)/(a + b)

= (45 x 40)/(45 + 40)

= 45 x 40/85

= 21.1764

= 22 min (approx).

Suppose B joined for x months . Then,  ( 85000 * 12 )/(42500 * x) = 3. or x = (85000 * 12) / (42500 * 3) = 8.

So, B joined for 8 months.

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=>ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

Speed =   6005×60 m/sec= 2 m/sec = 2 x (18/5) km/hr = 7.2 km/hr